The lengths of salamanders have a normal distribution with mean 15cm, and standard deviation 2cm. What length of salamander would place it at the 80th percentile of salamander lengths?

16.68 cm

Step-by-step explanation:

Given that :

Mean , m = 15

Standard deviation, s = 2

Length of salamander that would Place it at 80% of salamander length :

P(Z ≤ x) = 0.8

Zscore equivalent to 0.8 = 0.842

Using the relation :

Zscore = (x - m) /s

0.842 = (x - 15) / 2

1.684 = x - 15

Add 15 to both sides

1.684 + 15 = x - 15 + 15

16.684 = x

Hence, x = 16.68 cm

95.03 is the answer. hope this !

step-by-step explanation:

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